# 3x-y+3=0;(-1,2)write the standard form of the equation of the line that is parallel to the graph of the given equation and that passes through the point with the given coordinates.

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### 1 Answer

You should convert the standard form of the given equation to the slope intercept form `y = mx + n` , hence, you need to isolate y to the left side such that:

`-y = -3x - 3 => y = 3x + 3`

Comparing the result to the slope intercept form y = mx + n yields that m = 3 (slope) and n = 3 (y intercept).

You should remember that the slopes of two parallel lines have equal values, hence, the slope of the line whose equation will be found is also 3.

You need to write the point slope form of equation of a line such that:

`y - y_1 = m(x - x_1)`

Notice that the coordinates `x_1` and `y_1` are provided, hence, you need to substitute -1 and 2 for `x_1` and `y_1` such that:

`y -2 = 3(x + 1)`

You need to convert the point slope form into the standard form such that:

`y - 2 - 3x - 3 = 0`

`-3x + y - 5 = 0 =gt 3x - y + 5 = 0`

**Hence, evaluating the standard form of equation of the line, under the given conditions, yields `3x - y + 5 = 0` .**